`lim_(x to -1)f(x)` exists, find c given `f(x)={{:(x+2,x le -1),(cx^(2), x gt -1)}`

Answer

To find the value of \( c \) such that the limit \( \lim_{x \to -1} f(x) \) exists, we need to ensure that the right-hand limit (RHL) and left-hand limit (LHL) at \( x = -1 \) are equal.

The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} x + 2 & \text{if } x \leq -1 \\ c x^2 & \text{if } x > -1 \end{cases} \]

### Step 1: Calculate the Right-Hand Limit (RHL)

The right-hand limit as \( x \) approaches -1 is given by:
\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} c x^2 \]
Since \( x \) is approaching -1 from the right, we can substitute \( x = -1 + h \) where \( h \) approaches 0:
\[ \lim_{h \to 0} c(-1 + h)^2 = \lim_{h \to 0} c(1 - 2h + h^2) = c(1 - 0 + 0) = c \]
Thus, the right-hand limit is:
\[ \text{RHL} = c \]

### Step 2: Calculate the Left-Hand Limit (LHL)

The left-hand limit as \( x \) approaches -1 is given by:
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (x + 2) \]
Substituting \( x = -1 - h \) where \( h \) approaches 0:
\[ \lim_{h \to 0} (-1 - h + 2) = \lim_{h \to 0} (1 - h) = 1 \]
Thus, the left-hand limit is:
\[ \text{LHL} = 1 \]

### Step 3: Set RHL equal to LHL

For the limit \( \lim_{x \to -1} f(x) \) to exist, we need:
\[ \text{RHL} = \text{LHL} \]
This gives us:
\[ c = 1 \]

### Conclusion

The value of \( c \) that makes the limit exist is:
\[ \boxed{1} \]